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May you tell me if my "symbols" are correct? I am really suffering with this mathematical logic problem!

No sane witness would lie if his lying would implicate him in a crime. Therefore, if any witness implicated himself in a crime, then if all witnesses were sane, that witness did not lie.

This is my interpretation...

Premise:

(For all x) [(Lx implies Ix) implies [(Wx.Sx) implies (not Lx)]]

Conclusion:

(For all x) [(Wx and Sx) implies [(Wx implies Sx) implies (Wx implies (not Lx))]]

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The premise is fine, but the conclusion is off ... You need two quantifiers:

$\forall x ((Wx \land Ix) \rightarrow (\forall y (Wy \rightarrow Sy) \rightarrow \neg Lx))$

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  • $\begingroup$ Do I need two quantifiers because the problem says "that witness"? $\endgroup$
    – Beginner
    Dec 5 '16 at 18:45
  • $\begingroup$ May you explain me why do I need two quantifiers? Thank you! $\endgroup$
    – Beginner
    Dec 5 '16 at 18:48
  • $\begingroup$ The first quantifier is for the subject term: the witness 'x' whose lying would implicate himself. The second quantifier is for the antecedent in the predicate term, which states that 'if all witness are sane': since this is about 'all witness', we cannot simply use the 'x' again, but that would be saying 'if that particular witness 'x' is sane', which is a much weaker statement than 'all witness are sane'. $\endgroup$
    – Bram28
    Dec 5 '16 at 18:55
  • $\begingroup$ Amazing and very helpful explanation! Thank you! You are a genius! I am sure that you already knew it! $\endgroup$
    – Beginner
    Dec 5 '16 at 19:10
  • $\begingroup$ @Beginner You're welcome! But do everyone on this site a favor and learn how to typeset your questions in Latex. If you hit 'edit' you can see the source for any post .. you'll learn the basics real fast that way! $\endgroup$
    – Bram28
    Dec 5 '16 at 19:48

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